I tried a quick and dirty analytic geometry solution during lunch (essentially equivalent to Rachel's solution), and made some algebraic mistakes. I'm convinced there's a far more elegant geometric solution. Along this line (and admittedly working from the known answer), I have worked backwards through Heron's formula for the area of a triangle and shown that the area of the vertically aligned rectangle bounded by the centers of the two circles is the ratio of the base to the hypotenuse, when the length of the diagonal of that rectangle is defined as unity. This is just too pat an answer not to have a geometric meaning.
Here's the simplified calculations showing that to be the case:
If you let the radius of the small circle be a, then the radius of the large circle is Ka. The hypotenuse of the tinted rectangle is obviously a(K+1). As this is defined to be length 1, we have that a=1/(K+1). We can easily see that the base is a(K-1), or (K-1)/(K+1). The height can be calculated from the Pythagorean theorem to be 2√K/(K+1). Finally, the area of the rectangle is 2√K(K-1)/(K+1)
2, which matches the ratio of base to hypotenuse of the large triangle.
OBTW: image of this rectangle has been posted, and will be linked when available.
Edit:
link Though when I awoke this morning I realized that the area of the rectangle was simply sinθ/2 cosθ/2, so I missed by a factor of two and a reciprocal in any case. :sigh: